What are the units of the photon energy equation E = hv?

What are the units of the photon energy equation E = hv?

• A. Watts
• B. Joules
• C. Kelvins
• D. Electronvolts

Answer: (B)

The energy of a photon can be calculated using the equation \( E = hv \), where \( E \) represents the energy of the photon, \( h \) is Planck's constant, and \( v \) is the frequency of the photon. In this context, energy is measured in joules. Planck's constant \( h \) has units of joule-seconds (\( J \cdot s \)), and frequency \( v \) is measured in hertz (Hz), which is equivalent to per second (\( s^{-1} \)). When you multiply these two quantities together, you get the units for energy: \[ E = hv \implies [E] = (J \cdot s)(s^{-1}) = J \] Thus, the units of energy, \( E \), are joules. While electronvolts are a common unit for measuring energy in the context of subatomic particles, the standard SI unit of energy is the joule, making it the appropriate answer for this equation. Other options, such as watts (which are a measure of power, not energy) and kelvins (which measure temperature), do not apply in this context. Therefore, the correct identification

The energy of a photon can be calculated using the equation ( E = hv ), where ( E ) represents the energy of the photon, ( h ) is Planck's constant, and ( v ) is the frequency of the photon. In this context, energy is measured in joules.

Planck's constant ( h ) has units of joule-seconds (( J \cdot s )), and frequency ( v ) is measured in hertz (Hz), which is equivalent to per second (( s^{-1} )). When you multiply these two quantities together, you get the units for energy:

[

E = hv \implies [E] = (J \cdot s)(s^{-1}) = J

]

Thus, the units of energy, ( E ), are joules.

While electronvolts are a common unit for measuring energy in the context of subatomic particles, the standard SI unit of energy is the joule, making it the appropriate answer for this equation. Other options, such as watts (which are a measure of power, not energy) and kelvins (which measure temperature), do not apply in this context. Therefore, the correct identification